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%I #6 Mar 30 2012 17:36:28
%S 1,1,2,2,1,4,1,1,5,3,1,1,9,6,3,1,1,12,11,7,3,1,1,22,21,14,8,3,1,1,30,
%T 38,28,16,9,3,1,1,55,71,58,35,18,10,3,1,1,77,123,110,75,41,20,11,3,1,
%U 1,141,230,220,154,93,47,22,12,3,1,1,201,389,410,314,196,111,53,24,13,3,1,1
%N Triangle read by rows: T(n,k) is the number of length n left factors of Dyck paths having k triple-rises (triple-rise = three consecutive (1,1)-steps).
%C Row n>=2 contains n-1 entries.
%C Sum of entries in row n is binomial(n,floor(n/2))=A001405(n).
%C T(n,0)=A191786(n).
%C Sum(k*T(n,k),k>=0)=A191787(n).
%F G.f.: G=G(t,z) is given by G = C + z*C*(t*(G - 1 - z - z^2*G) + 1 + z + z^2*G), where C=C(t,z) is the solution of the equation z^2*(t+z^2-t*z^2)*C^2 - (1 - z^2 + t*z^2)*C + 1 = 0.
%e T(6,2)=3 because we have UD(U[UU)U], (U[UU)U]DD, and (U[UU)U]DU, where U=(1,1) and D=(1,-1) (the triple-rises are shown between parentheses).
%e Triangle starts:
%e 1;
%e 1;
%e 2;
%e 2,1;
%e 4,1,1;
%e 5,3,1,1;
%e 9,6,3,1,1;
%p eqC := z^2*(t+z^2-t*z^2)*C^2-(1-z^2+t*z^2)*C+1 = 0: C := RootOf(eqC, C): G := C*(1+z+z^2-t*z-t*z^2)/(1-t*z*C-z^3*C+t*z^3*C): Gser := simplify(series(G, z = 0, 15)): for n from 0 to 13 do P[n] := sort(coeff(Gser, z, n)) end do: 1; 1; for n from 2 to 13 do seq(coeff(P[n], t, k), k = 0 .. n-2) end do; # yields sequence in triangular form
%Y Cf. A001405, A191786, A191787
%K nonn,tabf
%O 0,3
%A _Emeric Deutsch_, Jun 18 2011
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